# Find the Sum of a Series

1. Oct 11, 2011

### BraedenP

1. The problem statement, all variables and given/known data

I am asked to find the sum of the series:
$$\sum_{n=2}^{\infty}n(n-1)x^n, |x|<1$$

2. Relevant equations

3. The attempt at a solution

We've learned a bit about Taylor series, so I could approximate this answer by taking and summing the first few derivatives. However, I suspect a more exact answer is desired.

The second idea I had was to change the sum so that is starts at n=0:

$$\sum_{n=0}^{\infty}(n+2)(n+1)x^{n+2}, |x|<1$$

and then try to rearrange it into the form of a geometric series for which I know a sum formula. However, I haven't had much luck with my rearrangement.

Am I missing something here?

2. Oct 11, 2011

### Staff: Mentor

Try expanding the series. If you do this, notice that there is a common factor that you can pull out. You might be able to recognize the remaining series as a derivative (or something) of a common series.

3. Oct 11, 2011

### BraedenP

Hmm.. I'm trying to figure out what you mean by expand.

Do you want me to:
- Expand the terms in the sum so that I get $n^2-nx^n$
- Expand the series for a few iterations (i.e. write it down for n=2,3,4, etc.)
or
- Perform a Taylor expansion on it where I use the formula $c_n=\frac{f^{(n)}(a)}{n!}$

None seem to reveal anything factorable to me.

4. Oct 11, 2011

### Staff: Mentor

Write the terms in the series without using the summation sign. Obviously, you can't write all of them, since it's an infinite series, but write the first two or three, then ..., and the general term, and ...
No, not this at all! The series already is a Taylor series (more specifically, a Maclaurin series.

5. Oct 11, 2011

### BraedenP

Yeah, this was option #2 in my list. I tried that, and found that only x was factorable, thus creating $n(n-1)x\cdot x^{n-1}$

And this is closer to the geometric form I was looking for, but I don't see how I can apply the geometric series sum formula to it in this form; I need to get rid of the extra 'x' somehow...

6. Oct 11, 2011

### Staff: Mentor

You can factor x2 out of every term. What do you get for the first three terms of the series? I'm thinking that you don't understand what I mean when I say to expand the series.

7. Oct 11, 2011

### SammyS

Staff Emeritus
You mean like: 3x2 + 6x3 + ... even more? ... even more ?

8. Oct 11, 2011

### BraedenP

I get $2x^2+6x^3+12x^4+...+n(n-1)x^n$

Oh, so I did make a mistake there. But factoring out x2 still doesn't illuminate anything for me. By doing that, I now have $n(n-1)x^2\cdot x^{n-2}$.

I still see no way to rewrite this as a geometric series. Sorry if I'm being a bother; but basically the only way we've been taught to sum an infinite series is to rewrite it as a geometric series and use the sum formula.

Unless my algebra is horrible this evening, I just can't see the connection.

9. Oct 11, 2011

### Staff: Mentor

That's not quite right.

You should have gotten
$2x^2+6x^3+12x^4+...+n(n-1)x^{n -2} + ...$
$=x^2(2*1 + 3*2*x + 4*3*x^2 + ... + n(n-1)x^{n -2} + ...)$

Look at the part in the parentheses. Hint: derivative.

10. Oct 11, 2011

### BraedenP

Oh! It's the derivative of $nx^{n-1}$ which I can evaluate as a geometric series. How do you just "see" this stuff in these equations?

So if I integrate it, the definite integral will equal the sum of the geometric series generated by integration.

Does this mean that if I multiply the sum of that geometric series by x2 I will have the answer to my question?

11. Oct 11, 2011

### Staff: Mentor

And further, $nx^{n-1}$ is the derivative of xn, right? So what you have in your general term is the 2nd derivative of xn, the general term in a geometric series.
Probably from a misspent youth, differentiating and integrating infinite series.
I think you have the basic idea, sort of, but you need to be very careful and write down step by step what you have, including my corrections/guidance above.

12. Oct 11, 2011

### BraedenP

Awesome! But now I'm stuck on one more thing... I can't actually find the sum of the geometric series for which I have the derivative, because the summation starts at 2. Can I just find the sum as if it started at 0 and then subtract those first two terms?

Then I'll have that my original sum equals $$x^2\cdot \left (\sum_{n=0}^{\infty}nx^{n-1}\right )-1$$

13. Oct 11, 2011

### Staff: Mentor

You're not getting all of what I'm saying.

d/dx(xn) = nxn-1

and
$$\frac{d^2}{dx^2}x^n = n(n - 1)x^{n - 2}$$

14. Oct 11, 2011

### BraedenP

I get that part, but what I don't get is how to convert the sum I get from that series back into what I have for the original.

So if I apply the sum formula to the derivative form ($nx^{n-1}$), using x as r and n as a in $\frac{a}{1-r}$ and find this sum from the derivative, can I just derive the result (and multiply by x2 since we factored it out) and have it equal the sum of my original series?

15. Oct 11, 2011

### Staff: Mentor

Isn't this what you have?
$$\sum_{n=2}^{\infty}n(n-1)x^n$$
$$= x^2 \frac{d^2}{dx^2}\left(\sum_{n=0}^{\infty}x^n\right)$$

16. Oct 12, 2011

### BraedenP

Yep, that's what I've got (finally). Is that correct?

17. Oct 12, 2011

### HallsofIvy

Staff Emeritus
Well, it isn't finished yet. Actually do the differentiation. What do you get?